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Open file with absolute path in java

I wrote this code to read the content of a file to a bytes array.
It works fine when path (given in the constructor) is relative. But I would like it to work in an absolute path instead. I looked up in java File class docs but got confused. How can I changed it to work with absolute path?
File file = new File(path);
byte[] bytesArray = new byte[(int) file.length()];
FileInputStream fis = new FileInputStream(file);
fis.read(bytesArray);
fis.close();

In your code here;
File file = new File(path);
Your path String variable just needs to be absolute instead of relative.
I don't see why it would not work. Did you try to update the path variable to absolute path of your file?

I think you can create an object from the Path interface using a relative path and getting that path using the Paths class using the static get method. once done you can get the absolute path from the created object and use it as a string if you prefer.
I hope I have helped you.
Path path1 = Paths.get("res/ficheroPrueba.txt");
File file = new File(path1.toAbsolutePath().toString());

Creating a file in linux server Java

i upload a csv file from client side and i want to create this file in server side.
Here is my function
public void uploadFile(FileUploadEvent e) throws IOException{
UploadedFile uploadedCsv=e.getFile();
String filePath="//ipAdress:/home/cg/Temp/input/ressource.csv";
byte[] bytes=null;
if(uploadedCsv != null){
bytes=uploadedCsv.getContents();
BufferedOutputStream stream = new BufferedOutputStream(new FileOutputStream(new File(filePath)));
String filename = FilenameUtils.getName(uploadedCsv.getFileName());
stream.write(bytes);
stream.close();
}
}
When I want to write the file I get this exception (No such file or directory)
SEVERE: java.io.FileNotFoundException: /ipAdress:/home/cg/Temp/input/ressource.csv (No such file or directory)
Knowing that the / home / cg / Temp / input path is created on the server.

Could you try:
String filePath="////ipAdress/home/cg/Temp/input/ressource.csv";
Instead of:
String filePath="//ipAdress:/home/cg/Temp/input/ressource.csv";
And this:
new File(new URI(filePath))
Instead of:
new File(filePath)
Or you can use jcif API How can I open a UNC path from Linux in Java?

I would use the <file>.mkdirs(); at one level above the file itself.
So do String filePath="//ipAdress:/home/cg/Temp/input
File directory = new File(filePath);
directory.mkdirs();
You can then make the file
File tempFile = new File(directory + "/ressource.csv);
Or a cleaner solution all around is just use Files.createTempFile(prefix, suffix) this will create a file in the temp directory of the system.

The reason that your code does not work is that you are trying to use a UNC pathname on Linux. Linux does not support UNC pathnames ... natively. They are a Windows-ism.
Here's your example
"//ipAdress:/home/cg/Temp/input/ressource.csv";
If you try to use that on Linux, the OS will look for a directory in the root directory of the file system. The directory it will look for will have the name ipaddress: ... noting that there is a colon in the directory name!
That will most likely fail ... because no directory with that name exists in the / directory.. And the exception message you are getting is consistent with this diagnosis.
If you are doing this because you are trying to push files out to other systems then you are going to do it some other way. For example:
Use NFS and mount the other system's file systems on the server.
Use a Java implementation of UNC names; e.g. How can I open a UNC path from Linux in Java?
(Which ever way you do it, there are security issues to consider!)
trying this new File(new URI(filePath)) instead of new File(filePath) i get this erreur. SEVERE: java.lang.IllegalArgumentException: URI is not absolute
It won't work. A UNC name is NOT a valid URL or URI.

I have found a solution for this problem, but it's not smart and still and it works
String fileName="ressource.csv";
File f = new File(System.getProperty("user.home") + "/Temp/input",fileName);
if (f.exists() && !f.canWrite())
throw new IOException("erreur " + f.getAbsolutePath());
if (!f.exists())
Files.createFile(f.toPath());
if (!f.isFile()) {
f.createNewFile(); // Exception here
} else {
f.setLastModified(System.currentTimeMillis());
}
Pending a more intelligent solution

How to read a file that i created inside my the same package?

This is a chunk of data I'd like to access by a method.
I'm doing the following to read my file:
String fileName = "file.txt"
InputStream inputStream = new FileInputStream(fileName);
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
My file.txt is in the same package, but I still get FileNotFoundException.
I didn't use a path url to point to the file because I thought since this it going to be an android application, hard-coding the path might not work when deployed... Please correct me if I am wrong. Thanks bunch!

This shows how to do that. https://stackoverflow.com/a/14377185/2801237
Also the 'package' your class is in has nothing to do with the 'path' where the file is being executed from. (two different concepts, 'package' = folder hierarchy of java source code files), 'path' = location on a filesystem of a specific file, your APK is being 'executed' in a particular place, and the location it writes a file is associated with that (I actually don't know where 'offhand' it writes by default, because I always get cache dir, or sd card root, etc.)

You may use:
InputStream inputStream = this.getClass().getResourceAsStream(fileName);

Programmatically uploading a file?

I'm developing a small program that uploads and downloads files from my box account.
I looked at the docs about uploading files and I found this code:
BoxFolder rootFolder = BoxFolder.getRootFolder(api);
FileInputStream stream = new FileInputStream("My File.txt");
rootFolder.uploadFile(stream, "My File.txt");
stream.close();
I don't really understand how it works. Where can I put the path to the file I want to upload? Or should I use different code?

The constructor for FileInputStream takes in a path to a file. The example in the documentation is uploading a file with the path "./My File.txt" relative to the current directory.
To make it a bit clearer, here's an example using a full path:
BoxFolder rootFolder = BoxFolder.getRootFolder(api);
FileInputStream stream = new FileInputStream("/path/to/My File.txt");
rootFolder.uploadFile(stream, "My File.txt");
stream.close();

Opening a resource file in a servlet on Openshift

I'm in troubles with opening file within my web-app. I tried it locally within Eclipse and it works fine but when I try to deploy it on Tomcat 6 on Openshift it doesn't find resource files for my web-app. There are some txt files in a ProjectFiles directory stored in WEB-INF directory; the code that locally opens file is
String nome_file = "C\:\\Users\\miKKo\\workspace\\fantacalcio_project\\WebContent\\WEB-INF\\ProjectFiles\\Risultati\\risultati_" + nome_lega + ".txt";
BufferedReader reader = new BufferedReader(new FileReader(nome_file));
I've pushed them within Git in the same repository (on server I renamed my project in "ROOT") and I've substituted string with this
String nome_file = this.getServletConfig().getServletContext().getContextPath()+"/WebContent/WEB-INF/ProjectFiles/Risultati/risultati_" + nome_lega + ".txt";
but it doesn't work. I've also tried with a context attribute
/var/lib/openshift/51c6178a5004467630000019/jbossews/work/Catalina/localhost/_/WEB-INF/ProjectFiles
but the thrown exception is always
java.io.FileNotFoundException: (#path) (No such file or directory)
What can I do for this?

Say your file is in the following location:
/WEB-INF/ProjectFiles/Risultati/risultat_text_file.txt
Then using:
String path = "/WEB-INF/ProjectFiles/Risultati/risultat_text_file.txt";
InputStream inputStream = new FileInputStream(this.getServletConfig().getServletContext().getRealPath(path));
Should work for you. Note that, ServletContext.getRealPath() return the real OS path corresponding to the given virtual path.
Edit:
If this doesn't work for your case, you really need to revisit your virtual path. You can manually check that does this file exist in the expected directory in the war file or you can log the output of the getRealPath() method to examine what's really going on! If necessary you can put "/" in your getRealPath() method and examine what is your application's root path.

Since I don't get application's root realpath, I resolved in this way:
String path="/WEB-INF/ProjectFiles/Risultati/risultati_test.txt";
InputStream inputStream = this.getServletConfig().getServletContext().getResourceAsStream(path);
BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream));
and now it works. By the way, I also found useful informations here
getResourceAsStream() vs FileInputStream